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A current of 5A is passing through a circular conducting wire of diameter 10cm as shown. (AB diameter and ‘O’ is the center)
The magnetic field at ‘O’

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$1 \times 10^{- 5} T,$ out of the plane

$2 \times 10^{- 5} T,$ out of the plane

$1 \times 10^{- 5} T,$ into the plane

$2 \times 10^{- 5} T,$ into the plane

answer is D.

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Detailed Solution

Field at ‘O’ due to semi-intinite current wires get added and field due to current in semi circular segments get cancelled.
$B = 2 (\frac{μ_{0} i}{2 π d})$

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